Question

Let y = y(x) be the solution of the differential equation, ${\displaystyle {(x^2+1)}^2\frac{\text{dy}}{\text{d}x}+2x(x^2+1)\text{y}}$ = 1, such that y(0) = 0. If ${\displaystyle \sqrt{\text{ay}}\left(1\right)=\frac{\pi }{32}}$ then the value of  'a' is ______.

Options

• ${\displaystyle \frac{1}{2}}$

• 1

• ${\displaystyle \frac{1}{16}}$

• ${\displaystyle \frac{1}{4}}$

Answer 1

Let y = y(x) be the solution of the differential equation, ${\displaystyle {(x^2+1)}^2\frac{\text{dy}}{\text{d}x}+2x(x^2+1)\text{y}}$ = 1, such that y(0) = 0. If ${\displaystyle \sqrt{\text{ay}}\left(1\right)=\frac{\pi }{32}}$ then the value of  'a' is ${\displaystyle \underline{\frac{1}{16}}}$.

Explanation:

${\displaystyle {(x^2+1)}^2\frac{\text{dy}}{\text{d}x}+2x(x^2+1)\text{y}}$ = 1

${\displaystyle \frac{\text{dy}}{\text{d}x}+\frac{2xy}{x^2+1}=\frac{1}{{(x^2+1)}^2}}$

From linear differential equation

P = ${\displaystyle \frac{2x}{x^2+1}}$ Q = ${\displaystyle \frac{1}{{(x^2+1)}^2}}$

${\displaystyle {\text{ye}}^{\int \frac{2x}{x^2+1}\text{d}x}=\int \frac{1}{{(x^2+1)}^2}{\text{e}}^{\int \frac{2x}{x^2+1}\text{d}x}\text{d}x+\text{C}}$

${\displaystyle {\text{ye}}^{\log (x^2+1)}=\int \frac{1}{{(x^2+1)}^2}{\text{e}}^{\log (x^2+1)}\text{d}x+\text{C}}$

y(x² + 1) = ${\displaystyle \int \frac{1}{x^2+1}\text{d}x+\text{C}}$

y(x² + 1) = tan^–1(x) + C

y(0) = 0  ...(Given)

⇒ C = 0

y(x² + 1) = tan^–1x

${\displaystyle \sqrt{\text{a}}\text{y}\left(1\right)=\frac{\pi }{32}}$ then a = ?

Put x = 1  ...(i)

y(1){1 + 1} = tan^–11

y(1) = ${\displaystyle \frac{\pi }{8}}$

Substituting in equation (i)

${\displaystyle \sqrt{\text{a}}\times \frac{\pi }{8}=\frac{\pi }{32}}$

a = ${\displaystyle \frac{1}{16}}$